Proposition 4.1.27CA3187A3AD4F

Proof. Take $R$ so large that $K\subset B$. Let $\varepsilon >0$, and consider the set $E_\varepsilon =\{ x\in K:U^{\mu _ K}(x)<V(K)-\varepsilon \} $, which is Borel since $U^{\mu _ K}$ is lower semicontinuous. If $C_ B^*(E_\varepsilon )>0$, then by theorem 4.1.23 $E_\varepsilon $ contains a compact set of positive capacity, which is not polar by theorem 4.1.17; that theorem provides a probability measure $\nu $ on this compact set with finite continuous potential, hence with finite energy. Then $I(\mu _ K,\nu )=\int U^{\mu _ K}\, \mathrm d\nu \leq V(K)-\varepsilon $, contradicting theorem 4.1.7. Thus $C_ B^*(E_\varepsilon )=0$ for every $\varepsilon >0$, and countable subadditivity (theorem 4.1.19) and corollary 4.1.24 show that $\{ x\in K:U^{\mu _ K}(x)<V(K)\} $ is polar. The finite-energy measure $\mu _ K$ gives zero mass to this set (theorem 4.1.17), so $U^{\mu _ K}\geq V(K)$ $\mu _ K$-almost everywhere. Since $\int U^{\mu _ K}\, \mathrm d\mu _ K=V(K)$, it follows that $U^{\mu _ K}=V(K)$ $\mu _ K$-almost everywhere.

Let $x\in \operatorname{Supp}\mu _ K$. Every neighborhood of $x$ has positive $\mu _ K$-measure and therefore contains points where $U^{\mu _ K}=V(K)$; by lower semicontinuity, $U^{\mu _ K}(x)\leq V(K)$. To extend this bound to $\mathbb R^ d$, decompose $\mu _ K=\sum _ j\mu _ j$ by theorem 4.1.10, which applies because $E_ d*\mu _ K$ is finite $\mu _ K$-almost everywhere, and let $\pi _ N=\sum _{j\leq N}\mu _ j$. The potential $U^{\pi _ N}$ is continuous on $\mathbb R^ d$ and, by differentiation under the integral sign, harmonic off the compact set $S_ N=\operatorname{Supp}\pi _ N\subset \operatorname{Supp}\mu _ K$. Since the kernel is nonnegative on $K\times K$, we have $U^{\pi _ N}\leq U^{\mu _ K}\leq V(K)$ on $S_ N$.

We claim that $U^{\pi _ N}\leq V(K)$ on $\mathbb R^ d$. If $\pi _ N=0$, this holds because $V(K)\geq 0$, so let $\pi _ N\neq 0$. As $|x|\to \infty $, the Newtonian potential $U^{\pi _ N}(x)$ tends to $0<V(K)$, and the logarithmic potential tends to $-\infty $. Applying the maximum principle (proposition 3.1.10) to $U^{\pi _ N}$ on the bounded open sets $B(0,\rho )\setminus S_ N$ with $\rho $ large, we obtain the claim. Finally, fix $x\in \mathbb R^ d$. The kernel $k(x,\cdot )$ is bounded below on $K$ by a constant $-c_ x$ (with $c_ x=0$ in the Newtonian case), and monotone convergence applied to $k(x,\cdot )+c_ x$ gives $U^{\pi _ N}(x)\to U^{\mu _ K}(x)$. Hence $U^{\mu _ K}(x)\leq V(K)$. $\square$